Syntax of command line arguments - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the main function with command line arguments.

C

int main(int [1], char *argv[]) {
    return 0;
}

Drag options to blanks, or click blank then click option'

Aargc

Bargv

Cargc[]

Dargs

Attempts:

3 left

💡 Hint

Common Mistakes

Using argv as the first parameter name instead of argc.

Adding brackets to argc like argc[].

The first parameter of main for command line arguments is argc, which counts the arguments.

2fill in blank

medium

Complete the code to access the first command line argument (excluding program name).

C

printf("First argument: %s\n", [1][1]);

Drag options to blanks, or click blank then click option'

Aargv[]

Bargc

Cargs

Dargv

Attempts:

3 left

💡 Hint

Common Mistakes

Using argc instead of argv to access arguments.

Trying to access argv with brackets like argv[].

The array argv holds the command line arguments as strings. Index 1 is the first argument after the program name.

3fill in blank

hard

Fix the error in the main function declaration for command line arguments.

C

int main([1] argc, char *argv[]) {
    return 0;
}

Drag options to blanks, or click blank then click option'

Afloat

Bvoid

Cint

Dchar

Attempts:

3 left

💡 Hint

Common Mistakes

Using void or char instead of int for argc.

Omitting the type before argc.

The first parameter argc must be of type int to hold the count of arguments.

4fill in blank

hard

Fill both blanks to declare main with command line arguments correctly.

C

int main([1] argc, [2] *argv[]) {
    return 0;
}

Drag options to blanks, or click blank then click option'

Aint

Bchar

Cfloat

Dvoid

Attempts:

3 left

💡 Hint

Common Mistakes

Using float or void types.

Swapping the types of the parameters.

The first parameter is int argc and the second is char *argv[] to hold argument strings.

5fill in blank

hard

Fill all three blanks to print all command line arguments using a loop.

C

for (int [1] = 0; [2] < [3]; [1]++) {
    printf("Arg %d: %s\n", [1], argv[[1]]);
}

Drag options to blanks, or click blank then click option'

Ai

Bargc

Ccount

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inconsistently.

Comparing to a variable other than argc.

Use i as the loop variable and compare it to argc to loop through all arguments.

Practice

1.

What does the argc parameter represent in a C program's main function?

easy

A. The number of command line arguments including the program name

B. The array of command line argument strings

C. The first command line argument

D. The program's exit status

Solution

1. Step 1: Understand the role of argc

   argc stands for 'argument count' and tells how many arguments are passed to the program including the program name itself.

2. Step 2: Differentiate argc from argv

   argv is the array holding the actual argument strings, not the count.

3. Final Answer:

   The number of command line arguments including the program name -> Option A

4. Quick Check:

   argc = argument count [OK]

Hint: Remember: argc counts all arguments including program name [OK]

Common Mistakes:

• Confusing argc with argv
• Thinking argc excludes the program name
• Assuming argc is the argument array

2.

Which of the following is the correct syntax for the main function to accept command line arguments in C?

easy

A. int main(char argc, char *argv[])

B. void main()

C. int main(int argc, char *argv[])

D. int main(int argv, char *argc[])

Solution

1. Step 1: Recall the standard signature for main with arguments

   The correct signature uses int argc and char *argv[] to receive argument count and argument strings.

2. Step 2: Identify incorrect options

   void main() lacks parameters, C uses wrong type for argc, D swaps argc and argv incorrectly.

3. Final Answer:

   int main(int argc, char *argv[]) -> Option C

4. Quick Check:

   Main signature with argc and argv = int main(int argc, char *argv[]) [OK]

Hint: argc is int, argv is array of char pointers [OK]

Common Mistakes:

• Using void main instead of int main
• Swapping argc and argv types
• Using char type for argc

3.

What will be the output of the following C program if run as ./program hello world?

int main(int argc, char *argv[]) {
    printf("%d %s %s", argc, argv[1], argv[2]);
    return 0;
}

medium

A. 3 hello world

B. 2 hello world

C. 3 ./program hello

D. Error: argv[2] does not exist

Solution

1. Step 1: Count the arguments passed

   Running ./program hello world passes three arguments: program name, "hello", and "world". So, argc = 3.

2. Step 2: Identify argv values

   argv[0] = "./program", argv[1] = "hello", argv[2] = "world".

3. Final Answer:

   3 hello world -> Option A

4. Quick Check:

   argc=3, argv[1]=hello, argv[2]=world [OK]

Hint: argv[0] is program name; argc counts all args [OK]

Common Mistakes:

• Assuming argc excludes program name
• Confusing argv indexes
• Thinking argv[2] is out of range

4.

Identify the error in this C program snippet that uses command line arguments:

int main(int argc, char *argv[]) {
    printf("%s", argv[argc]);
    return 0;
}

medium

A. argc should be a char type

B. Accessing argv[argc] causes out-of-bounds error

C. argv should be int array

D. Missing return statement

Solution

1. Step 1: Understand argv indexing

   argv indexes go from 0 to argc-1. Accessing argv[argc] is out of bounds and unsafe.

2. Step 2: Check other options

   argc must be int, argv is array of char pointers, and return statement is present.

3. Final Answer:

   Accessing argv[argc] causes out-of-bounds error -> Option B

4. Quick Check:

   argv valid indexes: 0 to argc-1 [OK]

Hint: argv max index is argc-1, not argc [OK]

Common Mistakes:

• Using argv[argc] instead of argv[argc-1]
• Changing argc type incorrectly
• Ignoring array bounds

5.

How would you write a C program that prints all command line arguments except the program name, each on a new line?

hard

A. Print argc without looping

B. Print argv[0] only

C. Use a loop from 0 to argc and print argv[i]

D. Use a loop from 1 to argc-1 and print argv[i]

Solution

1. Step 1: Understand argument indexing

   argv[0] is the program name, so to print only arguments, start from index 1.

2. Step 2: Loop through arguments

   Loop from i=1 to i<argc to print each argument on a new line.

3. Final Answer:

   Use a loop from 1 to argc-1 and print argv[i] -> Option D

4. Quick Check:

   Skip argv[0], loop i=1 to argc-1 [OK]

Hint: Start loop at 1 to skip program name [OK]

Common Mistakes:

• Starting loop at 0 printing program name
• Looping beyond argc-1 causing errors
• Not using a loop to print all arguments